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Multiplication of Equations

As we know, multiplication of integers may be defined in terms of addition. This is why Euclid did not mention multiplication among his Common Notions. Properties of multiplication are derivable from those of addition and, therefore, if we can add equations, we may multiply them as well. It's interesting that Euclid stipulates among Common Notions not only validity of addition but also that it's permissible to subtract equal quantities from equal quantities without disturbing their equality. The modern math would derive one property from another; however, the fact remains that both addition and its inverse operation (subtraction) may apply to equations. A legitimate question is whether the same is true of multiplication and the operations inverse to multiplication.

Division is one operation that reverses the result of multiplication. However, division has limitations: it's impossible to divide by 0. Other than that if a = b then a/r = b/r. However, concealed division by 0 results in errors and what's commonly known as arithmetic paradoxes. Here is a well known one: assume a = b and that both are positive numbers. Write

1.a,b>0that's given
2.a = bthat's also given
3.ab = b2this is step 2 times b
4.ab-a2 = b2-a2Euclid's Common Notion (3)
5.a(b-a) = (b+a)(b-a)distributive law
6.a = b+adividing by (b-a)
7.0 = bEuclid's Common Notion (3)
8.b = 2bEuclid's Common Notion (2)
9.1 = 2Nice, right?

What went wrong?


Another case of reversing the result of multiplication is squaring. Let a×a = b. Given b, what is a? As we know, a = b. As in the case of division, squaring should be used cautiously.

If two numbers are equal, their squares are also equal. Is the reverse true? That is, is it true that squares of two numbers are equal provided the numbers are equal to start with? In short, does u = v imply u = v? Or simpler yet: is the square root of a number unique?

Assume the answer is unqualified yes and let's see where this will lead us.

1. (n+1)2 = n2+2n+1Do not you know it?
2. (n+1)2-(2n+1) = n2Euclid's Common Notion (3)
3. (n+1)2-(2n+1)-n(2n+1) = n2-n(2n+1)ditto
4. (n+1)2-(n+1)(2n+1) = n2-n(2n+1)plain factoring
5. (n+1)2-(n+1)(2n+1)+(2n+1)2/4 = n2-n(2n+1)+(2n+1)2/4Euclid's Common Notion (2)
6. [(n+1)-(2n+1)/2]2 = [n-(2n+1)/2]2a little algebra
7. (n+1)-(2n+1)/2 = n-(2n+1)/2taking square roots of both sides
8. n+1 = nEuclid's Common Notion (3)
9. 1 = 0That's also nice, right?

What went wrong?

References

  1. T. Pappas, The Joy of Mathematics, Wide World Publishing, 1989
  2. J. R. Newman, The World of Mathematics, v3, Dover, 2003

Copyright © 1996-2008 Alexander Bogomolny

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